Associate Certified Electronics Technician (CET) Practice Exam 2025 - Free CET Practice Questions and Study Guide

Question: 1 / 400

Three capacitors with values of 470 pf, 560 pf, and 0.47 pf are connected in parallel. What is the total capacitance?

1120 pf

1030.47 pf

When capacitors are connected in parallel, the total capacitance is simply the sum of the individual capacitances. This is because each capacitor adds its capacity to store charge independently of the others.

In this case, the three capacitors have values of 470 picofarads (pf), 560 picofarads (pf), and 0.47 picofarads (pf). To find the total capacitance, you need to first make sure all values are expressed in the same unit. Since 0.47 pf is equivalent to 470 femtofarads (which isn't significant compared to the others), it can be added directly to the other capacitances.

Now, you sum the values:

- 470 pf

- 560 pf

- 0.47 pf (or can be treated as just 0.47 pf)

Performing the calculation:

Total capacitance = 470 pf + 560 pf + 0.47 pf = 1030.47 pf.

This calculation confirms that the total capacitance of the system of capacitors in parallel is indeed 1030.47 picofarads.

The option that correctly reflects this total capacitance is B, making it the correct answer for this question.

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1.6 nF

560 pf

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